Linear Equations Reference

Equations of Lines

A line is determined by its slope and one point, or by two points. Coordinate geometry also asks us to measure segments, find midpoints, and build perpendicular bisectors.

Solved Examples

Jump directly to a worked example by problem type.

Fact Table

Given Best Form What to Do
Slope and y-intercept. $y=mx+b$ Insert $m$ and $b$ directly.
Slope and one point. $y=mx+b$ or point-slope form. Insert the point for $(x,y)$ and solve for $b$.
Two points. Start with the slope formula. Find $m$, then use either point to solve for $b$.
Horizontal line. $y=c$ The slope is $0$.
Vertical line. $x=c$ The slope is undefined, so it is not $y=mx+b$.

Content Formulas

Slope-Intercept Form
$$y=mx+b$$
Point-Slope Form
$$y-y_1=m(x-x_1)$$
Slope Formula
$$m=\frac{y_2-y_1}{x_2-x_1}$$
Parallel and Perpendicular Slopes
$$m_{\parallel}=m$$ $$m_{\perp}=-\frac1m$$
A point like $(4,-3)$ means $x=4$ and $y=-3$. When using $y=mx+b$, substitute both coordinates and the slope before solving for $b$.

Classic Examples

Point and Slope

Find an equation of the line that passes through $(4,-3)$ and has slope $2$.

Solution Steps
  1. Write slope-intercept form, $y=mx+b$.
  2. Substitute the given slope and point coordinates.
  3. Simplify the substitution.
  4. Solve for the y-intercept $b$.
  5. State the equation of the line: $y=2x-11$.
$$y=mx+b$$ $$-3=2(4)+b$$ $$-3=8+b$$ $$b=-11$$ $$y=2x-11$$

Point-Slope First

Write the line through $(-2,5)$ with slope $-\frac32$.

Solution Steps
  1. Write point-slope form.
  2. Substitute the slope and point coordinates.
  3. Simplify the grouped expression.
  4. Distribute and isolate $y$.
  5. State the equation: $y=-\frac32x+2$.
$$y-y_1=m(x-x_1)$$ $$y-5=-\frac32(x-(-2))$$ $$y-5=-\frac32(x+2)$$ $$y=-\frac32x-3+5$$ $$y=-\frac32x+2$$

Two Points

Find the equation of the line through $(1,4)$ and $(5,12)$.

Solution Steps
  1. Use the two points in the slope formula.
  2. Evaluate the slope.
  3. Substitute one point and the slope into $y=mx+b$.
  4. Solve for $b$.
  5. State the equation: $y=2x+2$.
$$m=\frac{12-4}{5-1}$$ $$m=2$$ $$4=2(1)+b$$ $$b=2$$ $$y=2x+2$$

Parallel and Perpendicular Lines

A line has slope $-\frac23$. Find the slope of a parallel line and a perpendicular line.

Solution Steps
  1. Keep the same slope for a parallel line and state $m_{\parallel}=-\frac23$.
  2. Take the negative reciprocal and state $m_{\perp}=\frac32$.
$$m_{\parallel}=-\frac23$$ $$m_{\perp}=\frac32$$

Coordinate Segment Table

Task Use What It Means
Find distance or length. Distance formula. Measures the length of the segment between two points.
Find the midpoint. Average the x-coordinates and average the y-coordinates. Gives the point halfway between the endpoints.
Find a perpendicular bisector. Midpoint plus perpendicular slope. The line crosses the segment at its midpoint and makes a right angle.
Horizontal segment. Perpendicular bisector is vertical. The bisector has equation $x=c$.
Vertical segment. Perpendicular bisector is horizontal. The bisector has equation $y=c$.

Coordinate Segment Formulas

Distance / Segment Length
$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$
Midpoint
$$M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$$
Perpendicular Slope
$$m_{\perp}=-\frac1m$$
Perpendicular Bisector
$$\text{line through midpoint with slope }m_{\perp}$$
A perpendicular bisector is not just perpendicular. It must also pass through the midpoint of the segment.

Coordinate Segment Examples

Distance and Length

Find the length of the segment from $A(2,-1)$ to $B(8,7)$.

Solution Steps
  1. Substitute the coordinates into the distance formula.
  2. Evaluate the coordinate differences and their squares.
  3. Evaluate the square root.
  4. State the segment length: $10$.
$$d=\sqrt{(8-2)^2+(7-(-1))^2}$$ $$=\sqrt{6^2+8^2}$$ $$=\sqrt{100}$$ $$=10$$

Midpoint

Find the midpoint of $A(-4,6)$ and $B(10,-2)$.

Solution Steps
  1. Average the x-coordinates.
  2. Average the y-coordinates.
  3. State the midpoint: $(3,2)$.
$$M=\left(\frac{-4+10}{2},\frac{6+(-2)}{2}\right)$$ $$=\left(\frac62,\frac42\right)$$ $$=(3,2)$$

Perpendicular Bisector

Find the perpendicular bisector of the segment with endpoints $A(1,2)$ and $B(7,4)$.

Solution Steps
  1. Find the midpoint of the segment.
  2. Find the slope of the segment.
  3. Simplify the segment slope.
  4. Take its negative reciprocal.
  5. Use point-slope form through the midpoint.
  6. Distribute and isolate $y$.
  7. State the perpendicular bisector: $y=-3x+15$.
$$M=\left(\frac{1+7}{2},\frac{2+4}{2}\right)$$ $$=(4,3)$$ $$m_{AB}=\frac{4-2}{7-1}$$ $$=\frac13$$ $$m_{\perp}=-3$$ $$y-3=-3(x-4)$$ $$y=-3x+15$$